When chlorine is passed into boiling toluene in the presence of light until the increase in weight corresponds to roughly two chlorine atoms, a product (A) is obtained. When (A) is hydrolyzed with \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution, it is converted to (B). The product (A) and (B) are respectively

Short Answer

Expert verified
Answer: Product (A) is ortho-dichlorotoluene (1,2-dichloro-4-methylbenzene), and product (B) is o-dihydroxytoluene (1,2-dihydroxy-4-methylbenzene).

Step by step solution

01

Identify the reactants

Toluene (C\(_7\)H\(_8\)) is the initial reactant for the first reaction, and chlorine (Cl\(_2\)) is added. The increase in weight is roughly equivalent to two chlorine atoms, which means two chlorine atoms are being added to the toluene molecule.
02

Chlorination of toluene

Toluene undergoes chlorination in the presence of light. Since the increase in weight is roughly equivalent to two chlorine atoms, two hydrogen atoms from the toluene molecule are being replaced with two chlorine atoms. This leads to the formation of product (A), which is ortho-dichlorotoluene (1,2-dichloro-4-methylbenzene).
03

Hydrolysis of ortho-dichlorotoluene

In the second reaction, product (A) undergoes hydrolysis with calcium hydroxide (\(\mathrm{Ca}(\mathrm{OH})_{2}\)) solution. This leads to (B) being formed. The hydrolysis replaces the two chlorine atoms with two hydroxide (OH) groups on the benzene ring, forming o-dihydroxytoluene (1,2-dihydroxy-4-methylbenzene).
04

Final products

Product (A) is ortho-dichlorotoluene (1,2-dichloro-4-methylbenzene) and product (B) is o-dihydroxytoluene (1,2-dihydroxy-4-methylbenzene) respectively.

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